Universal logarithmic formulas for Segre classes on curve Quot schemes
Determine whether the generating series of Segre pushforwards for tautological bundles on relative curve Quot schemes admits a universal logarithmic expression in the relative classes \(\kappa_{a,b}=\pi_\star(c_1(L)^a c_1(\omega_\pi)^b)\), and characterize its universal power series by a closed recursion extending the recursion for relative symmetric products.
References
Motivated by their study of Quot schemes, we ask the following question.
For $N\ge1$, let $\pi:C\to B$ be a smooth projective curve family, let $L$ be a line bundle on $C$, and let $\rho_n:\mathfrak Q_n=\operatorname{Quot}{C/B}(O_C{\oplus N},n)\to B$. Denote by $L{\mathfrak Q}{[n]}$ the tautological bundle on $\mathfrak Q_n$. Does
\sum_{n\ge0}qn(\rho_n)_\star s\bigl(L_{\mathfrak Q}{[n]}\bigr)
admit a universal logarithmic formula in the classes $\kappa_{a,b}=\pi_\star(c_1(L)a c_1(\omega_\pi)b)$? Can its universal series be characterized by a closed recursion extending Theorem \ref{segre_curve}?
It is natural to ask whether this open-locus modular interpretation admits a compactified extension.
For each $n$, let $\overline{\tau}n: [\overline{\mathcal M}{g,A_n}/S_n] \longrightarrow \overline{\mathcal M}g$ be the morphism forgetting the weighted markings and stabilizing. Let $\overline p_n:\overline{\mathcal U}_n \longrightarrow [\overline{\mathcal M}{g,A_n}/S_n]$ be the universal curve, let $\overline{\mathcal D}n\subset\overline{\mathcal U}_n$ be the universal unordered divisor of the weighted markings, and set $\overline{\mathbb V}_n=(\overline p_n)\star O_{\overline{\mathcal D}_n}$. This defines a compactified Segre series
$\overline{\mathsf Z}{\mathrm{Has}_g(q)
\sum_{n\ge0}qn\,(\overline{\tau}n)\star s(\overline{\mathbb V}_n) \in A*(\overline{\mathcal M}_g)_Q[[q]]$.
Can the compactified series
\overline{\mathsf Z}{\mathrm{Has}_g(q)
\sum_{n\ge0}qn\,(\overline{\tau}n)\star s(\overline{\mathbb V}_n)
be computed concretely in the tautological ring $R*(\overline{\mathcal M}_g)_Q[[q]]$?